Check out this c++ solution

[raju-gupta_839]

int found_or_not = -1;
int dfs(TreeNode* A, int B, int C){
    if(found_or_not > 0){  // if lca is found
        return 0;
    }
    if(!A){  // NULL case
        return 0;
    }
    
    if(A->val == B && A->val == C){  // lca is A->val if "B == C == A->val"
        found_or_not = A->val;
        return 0;
    }
    
    int l = dfs(A->left, B, C);
    int r = dfs(A->right, B, C);
    
    if(A->val == B || A->val == C){   // if lca is any one of B || C
        if(l || r){ // one among B |`Preformatted text`| C is already detected
            found_or_not = A->val;
            return 0;
        }
        return 1;
    }if(l & r){  // if lca is not any one of B || C
        found_or_not = A->val;
        return 0;
    }
    return l || r;  // any one value found or not case
}
 
int Solution::lca(TreeNode* A, int B, int C) {
    found_or_not = -1;  // reset value for multiple test cases
    dfs(A, B, C);
    return found_or_not;
}